High-speed railway operation plan considering passenger demand under public health events

doi:10.16265/j.cnki.issn1003-3033.2023.06.1410

Abstract

Abstract:

To optimize high-speed railway operation plan under public events, a multi-objective optimization model was established with the highest expectations of infectious disease prevention and control, the lowest interchange fees in public health events and the lowest enterprise operation cost as the optimization objective. It took the carrying capacity and train capacity under the infectious disease prevention and control policy as the constraints. The TSABC algorithm was designed to solve the model. Matlab programming was utilized to verify the rationality of the model and algorithm with the example of Wuhan-Guangzhou high-speed railway. The results indicate that compared with the artificial bee colony algorithm, the improved algorithm has better solution quality. Compared with the initial solution, the optimal solution has reduced by 28.90% in the prevention and control penalty value. Compared to the situation of non-infectious disease prevention and control, it is reduced by 40.53% and the total number of running trains decreased by 21. The result shows that during the prevention and control of public health events, the operation plan can be optimized by reducing the number of trains and stops at high-risk stations, and stopping at stations of the same risk level.

Key words: public health events, passenger demand, high speed railway, operation plan, infectious disease prevention and control

WANG Yue, MENG Xuelei, QIN Yongsheng, LI Yi'na. High-speed railway operation plan considering passenger demand under public health events[J]. China Safety Science Journal, 2023, 33(6): 166-173.

Figures/Tables 11

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Tab.1

Penalty values for prevention and control of infectious disease across various risk levels

始发站风险类型	途径站风险类型及疫情风险惩罚值 $α n$
高风险	中风险 $α 1$ =0.25	低风险 $α 2$ =0.75
中风险	高风险 $α 3$ =0.65	低风险 $α 4$ =0.35
低风险	高风险 $α 5$ =0.80	中风险 $α 6$ =0.20

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参数	取值	参数	取值	参数	取值
u_k/(元·km^-1)	170	σ/(元· 人^-1·min^-1)	25	λ₁	0.6
c_t/元	100	t_p/min	35	λ₂	0.2
N_s	8	I_r/min	10	λ₃	0.2

求解结果	Z	Z₁	Z₂	Z₃	列车数
正常情况	3 929 240	213.9	1 296 220 0	267 000	65
初始解	3 615 985	178.9	1 243 830 0	253 625	59
最优解	1 952 793	127.2	9 128 740	253 625	44

算法	Z	Z₁	Z₂	Z₃	列车数
文中算法	1 952 793	127.2	9 128 740	253 625	44
ABC算法	1 976 075.6	134.6	9 215 831	260 747	49

列车编号	停靠站点	编组类型	列车编号	停靠站点	编组类型	列车编号	停靠站点	编组类型	列车编号	停靠站点	编组类型
1	v₁、v₄、v₇、v₉、 v₁₀、v₁₂、v₁₅	1	12	v₁、v₂、v₄、v₆、v₇、 v₉、v₁₁、v₁₃、v₁₅	2	23	v₁、v₄、v₆、 v₁₁、v₁₂、v₁₅	2	34	v₁、v₄、v₈、 v₁₁、v₁₄、v₁₅	2
2	v₁、v₁₀、v₁₅	2	13	v₁、v₆、v₁₅	1	24	v₁、v₂、v₆、 v₉、v₁₂、v₁₅	1	35	v₁、v₄、v₆、 v₉、v₁₅	1
3	v₁、v₃、v₄、v₉、 v₁₁、v₁₂、v₁₅	1	14	v₁、v₄、v₆、v₈、 v₁₁、v₁₂、v₁₅	1	25	v₁、v₂、v₆、v₉、 v₁₁、v₁₂、v₁₅	1	36	v₁、v₄、v₇、 v₁₁、v₁₅	2
4	v₁、v₃、v₅、 v₆、v₁₅	2	15	v₁、v₂、v₆、 v₉、v₁₂、v₁₅	1	26	v₁、v₄、v₇、 v₁₀、v₁₅	1	37	v₁、v₆、 v₉、v₁₅	1
5	v₁、v₂、v₄、 v₇、v₁₁、v₁₄、v₁₅	1	16	v₁、v₃、v₆、v₇、 v₉、v₁₂、v₁₅	1	27	v₁、v₂、v₄、v₆、 v₉、v₁₁、v₁₅	1	38	v₁、v₄、v₇、 v₉、v₁₁、v₁₅	1
6	v₁、v₂、v₃、v₅、 v₇、v₉、v₁₁、v₁₅	2	17	v₁、v₆、v₉、v₁₅	1	28	v₁、v₂、v₆、 v₉、v₁₂、v₁₅	1	39	v₁、v₆、v₁₀、 v₁₁、v₁₃、v₁₅	2
7	v₁、v₆、v₉、v₁₅	1	18	v₁、v₃、v₄、v₇、 v₉、v₁₂、v₁₅	2	29	v₁、v₆、v₁₅	1	40	v₁、v₅、v₆、v₉、 v₁₂、v₁₄、v₁₅	1
8	v₁、v₂、v₆、 v₉、v₁₂、v₁₅	1	19	v₁、v₃、v₇、v₉、 v₁₀、v₁₄、v₁₅	1	30	v₁、v₆、v₁₁、 v₁₃、v₁₅	1	41	v₁、v₄、v₇、 v₁₀、v₁₃、v₁₅	1
9	v₁、v₃、v₄、v₆、 v₉、v₁₁、v₁₂、v₁₅	1	20	v₁、v₄、v₅、v₉、 v₁₂、v₁₄、v₁₅	2	31	v₁、v₂、v₄、v₆、 v₇、v₁₁、v₁₅	2	42	v₁、v₂、v₄、v₅、 v₉、v₁₁、v₁₂、v₁₅	1
10	v₁、v₆、v₁₅	1	21	v₁、v₂、v₄、v₈、 v₁₁、v₁₂、v₁₅	1	32	v₁、v₄、v₉、 v₁₁、v₁₂、v₁₅	1	43	v₁、v₄、v₆、 v₈、v₁₁、v₁₅	1
11	v₁、v₂、v₄、v₉、 v₁₁、v₁₃、v₁₅	1	22	v₁、v₆、v₁₁、v₁₅	1	33	v₁、v₆、v₉、 v₁₁、v₁₂、v₁₅	2	44	v₁、v₂、v₆、v₉、 v₁₂、v₁₅	1